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Estimating piston-engine thrust on take-off
In my Uni course I've learnt how a piston-engine produces a constant power, as opposed to a constant thrust like in a jet engine, which can be estimated as Thrust = Power / Velocity. However, using this formula I calculated that a Spitfire should be able to go from idle to take-off in under 1.5s (as opposed to 9.5s irl). So my question is this: what else is going on that makes the true take-off time so much longer than my calculated take-off time?

What does your formula give as thrust a 0 velocity? Or near zero... is that a realistic value?

So my question is this: what else is going on that makes the true take-off time so much longer than my calculated take-off time?

Findings: If evaluated considering propeller thrust, resisting drag, and weight of the aircraft, an accurate and realistic time and distance for reaching take-off velocity can be determined.  An adequately developed computational procedure must be used, however.  Aspects considered include propeller diameter and blade dimensions for the estimation of thrust, and influence of the propeller slip-stream on lift.  Results for two Spitfire aircraft, for which adequate factual information was available, show computed times and distances to takeoff that are comparable, with little difference, to reported field measurements. For aircraft K.9793, a well documented Mark 1 machine, the computed take-off-run time and distance was about 12.39s at 321 yds, while the field reported time and distance estimate was 12.5s at 320 yds, given no wind and standard atmosphere conditions.  An incidental investigation regarding computational adjustments to determine take-off time and distance in wind provided similarly comparable results.  A brief overview of the basic computational process for take-off time and distance without wind is provided. $^*$A short end-note gives a perspective of the time required to answer this question.

Your very interesting question involves somewhat more than meets the eye, particularly when trying to make an assessment of a plane of this type.  A direct solution can be determined only if a number of variously different factors are considered.  You see, in order to obtain realistic results to determine the take-off time and distance for the Spitfire, one must, in a literal sense, fly the airplane on paper.  This means accounting for various factors of significance that affect the outcome of the determined result.
Three basic factors are at play, namely, thrust, drag, and weight.  From the outset, we would have to consider these three factors to determine the required take-off time and distance of a Spitfire.   Specific issues in accelerating an aircraft to take-off velocity primarily regard propeller thrust, drag due to rolling resistance during the take-off run, aerodynamic drag of the fuselage and lift-induced drag of the wing, and ultimately, weight of the aircraft and consequent lift produced by the wing during the take-off run.  In determining our answer, by assuming the process-forces are linear with respect to the square of the velocity of the aircraft, we only need to consider the mass of the plane, velocity at takeoff, and mean force of acceleration thru the take-off run.
Let’s take a closer look.  During takeoff, aerodynamic drag, lift-induced drag, and rolling resistance comprise forces opposing the accelerating thrust of the propeller.  During the take-off run, lift of the wing reduces rolling resistance by taking weight off the landing gear.  However, as the aircraft is accelerated, lift from the wing also creates drag and the force of this drag increases as speed of the aircraft, and hence force of lift, increases.  Included in resisting forces is aerodynamic drag of the aircraft itself (i.e landing gear, canopy, antenna mast), as well as drag caused by portions of the aircraft in the slip-stream flow of the propeller.  So, for any plane to take off, especially in still air, the force of acceleration must, at the start, move the aircraft against resisting forces that are due to the initiation of lift and dead weight (mass inertia) of the aircraft itself, drag caused by resistance of wheel bearings and landing-gear wheels rolling on the ground, aerodynamic drag caused by the aircraft fuselage and other appurtenances such as cowl vents, induction intake, and engine-cooling inlet.  All of these opposing forces must be overcome by thrust of the propeller. That thrust is developed through rotation of the propeller by the aircraft engine.
There are also several unknown factors.  For instance, given the presumed surface condition of the runway, and still air, we could assume minimal rolling resistance and optimum conditions for aircraft performance upon which to base an analysis.  With little information regarding piloting practice with this airplane, we have to presume the most conservative practice in operation of the aircraft, and use that as a starting point.  And we need additional information about the airplane and its propeller.  Some historical data and relevant information are available, however.  Most of the field test data for Spitfire take-off measurements, among other things, were measured at grass airfields near sea-level.  Data during these tests regarding wind, temperature, and pressure, were noted.  A host of specific information about the aircraft being tested was carried in the test reports.  These data include engine type, propeller blade dimensions, and all-up weight of the aircraft tested.  Exactly the information needed upon which to make an assessment of aircraft take-off performance.  With this information, the propulsive force during takeoff can be estimated, as can the time and distance required for takeoff.  These estimates can be compared with field measurements.
But there are additional, basic primary considerations: This aircraft has a tail wheel.  Consequently, as this aircraft is accelerated to take off, the tail is raised while keeping the nose slightly elevated.   In attempting to determine the most reasonably short time and distance to takeoff, one would presume the aircraft would leave the ground at airspeeds just above stall velocity.  For instance, a rather basic standard procedure in computing take-off airspeed for an aircraft of this type, is to a) assume a representative take-off lift coefficient at stall and determine the consequent stall velocity, then, b) multiply that velocity by a factor of 1.2 to assure sufficient airspeed to be in the positive command region of the power curve as the aircraft achieves flight velocity.  This assures that as the pilot folds the landing gear and the aircraft leaves ground-effect, sufficient airspeed is available to maintain control.  However, on takeoff, as the aircraft is near its lowest possible flight velocity and throttled at full power, the propeller slip-stream will have a significant influence on lift.   Consequently, to examine the Spitfire time-to-takeoff problem given a general approach and relevant factors, as just previously discussed, we must also consider the propeller slip stream and its influence on lift.    And finally there is aircraft pitch.  In addition to the propeller slip stream, we must account for the attitude of the wing in generating lift at takeoff.  By adjusting pitch, take-off velocity can be closely determined so that calculated take-off time and distance can be adequately compared with actual field measurements.   Essentially, this total process gives us a reality-check on our results.
The propeller is, perhaps, the most significant element in aircraft performance regarding this airplane.  Spitfire aircraft were operated on takeoff by advancing the throttle to rated take-off power at full boost.  As this occurs, the propeller generates tremendous thrust by inducing airflow through the propeller disk. As aircraft motion is initiated at the start of the take-off run, the force of this thrust can well exceed 2400 lbs.  As this flow passes behind the propeller, the diameter of the induced flow is contracted and the flow velocity is increased by a factor of about two.  This accelerated flow is the propeller slip stream, which then flows along and over the fuselage, and consequently over the near-fuselage, inner portion of the wing.  A resulting favorable consequence is that this higher-velocity slip-stream flow augments lift of the wing.
So what happens as the aircraft begins to move?  With full power, thrust of the propeller, the accelerating force moving the airplane, is at maximum.  This thrust, at zero forward velocity, is called static thrust.  Just as the plane begins to move, thrust becomes no longer static yet propeller efficiency increases.  Once in motion, available thrust from the propeller is consumed by opposing resisting forces due to rolling resistance of the aircraft as the run begins, and ultimately just aerodynamic drag, induced drag, and lift remain as the aircraft achieves flight.   The sum of these resisting forces over the course of the take-off run is taken to reduce available thrust of the propeller in linear relation to the square of the velocity of the aircraft. Consequently, one may calculate the accelerating force available from the propeller as a linear function of the square of the velocity of the aircraft.   The aircraft, of course, will achieve flight at the moment that lift is equal to weight of the aircraft.  Nevertheless, in order to determine the time and distance this will incur, we must know the aircraft weight (of course) and landing gear rolling resistance at the start of the run, area of the wing, slip-stream velocity due to the propeller, the resulting effective lift coefficient being developed by the wing, and ultimately, velocity and drag of the aircraft at takeoff.
And now a word from our sponsor –

Suppose we know the velocity at which an aircraft will take off, and the thrust – the force of acceleration – necessary to accomplish this, then what is the time and distance necessary for the aircraft to achieve this velocity? $  $ Let’s look at the math…
We work with variables $m$ = mass, $F$ = force, $t$ = time, $V$ = velocity, and $S$ = distance.  In this case, we take $F$ to be the mean force thru the path traveled.  If we already know $m$, and forces opposing $F$ are linear with regard to $V^2$, then these aspects follow:
$$F = m \cdot a$$
$$a = \frac{F}{m}$$
$$a \cdot t = V$$
$$V \cdot \frac{t}{2} = S$$
$$so \       \ V = \frac{2S}{t}$$
$$but \      \ t = \frac{V}{a}$$
$$therefore \     \ t =  \frac {m \cdot V}{F} .$$
Consequently, with already given mass $m$, if we know the take-off velocity $V$, and mean accelerating force $F$, then we can determine the time to takeoff.  Now, however,
$$given \     \ a = \frac{F}{m}$$
$$then \     \ V^2 = 2\ \ \frac{F}{m}\ S $$
$$reworking, we find \     \ S = \frac{m V^2}{2 F}  .$$
Consequently, again, with given mass $m$, by just knowing the take-off velocity $V$ and mean accelerating force $F$, we can determine also the distance required for the take-off run.

Given what we need to know, how was Spitfire take-off time and distance calculated?  In the abstract, one must calculate the take-off velocity first, and then determine the time and distance needed for available thrust to achieve that velocity, presuming available thrust, opposing drag, and consequent acceleration, are linear in relation to the square of velocity.  Available thrust is estimated as being equal to static thrust less opposing drag.  For the question at hand, the calculations were based on an empirically based understanding of propeller thrust, an accounting of the slip-stream influence on lift just as the aircraft achieves flight velocity, and conceptually direct aspects regarding drag, acceleration, velocity, and distance.  Resulting time and distance are dependent on a determination of the mean force of acceleration thru the takeoff run.  In short, all we need to know to determine take-off time and distance, is the velocity of the aircraft at takeoff, and mean thrust (force of acceleration) during the takeoff run.   That mean force is determined in relation to the thrust at the start of the take-off run, and thrust as the aircraft achieves flight.
Let’s take a quick descriptive overview of the computational accounting of thrust, drag resistance, lift, and velocity at takeoff.  Remember, the aircraft is very near or at its stalling velocity at takeoff save for the higher-velocity, propeller-accelerated slip-stream flow over the wing. The significance of the results will be summarized.  Here is the process –

Calculate the estimated stalling velocity at takeoff using a simple estimate of maximum $C_L$ at stall.  Adjust $V_{stall}$ by a factor of 1.2 to obtain estimated take-off velocity, $V_{t.o.}$.  Derive the estimated $C_L$ at takeoff from the estimated $V_{t.o.}$.
Calculate propeller static thrust and slip-stream velocity.  Then, for estimated dimensions of the fuselage frontal area, determine the slipstream diameter and region of the slipstream flowing over each wing.
The revised $C_L$ allows us to calculate the effective $C_L$ due to the slip stream. This requires knowing the specific wing chord and span dimensions, and consequent free-stream velocities relative to the wing, both interior to and exterior of the slip stream. We will also need to know the aircraft pitch and resulting angle of attack of the wing for each of the wing areas relative to the slip stream.  For the total wing, the effective $C_L$ and $V_{t.o.}$ is now determined.
Using the aircraft polar, calculate the force of drag on the aircraft at takeoff velocity.  Also, determine the rolling resistance force of the aircraft at the moment the take-off roll begins.  The mean thrust of acceleration due to the propeller is determined at the start of the run, and end of the run, using these values.  The mean force of acceleration is determined as $$F = \frac{F_s – F_e}{ln (\frac{F_s} {F_e})}$$ where $F_s$ is static thrust less rolling resistance starting the run, and $F_e$ is static thrust less total aerodynamic drag just as flight velocity $V_{t.o.}$ is achieved.
Determine consequent time and distance for takeoff using the mean force of acceleration during the run, and the determined velocity at takeoff.  Note: The mean force of acceleration is naturally biased toward the end of the take-off run.  Consequently, within limits, errors in the initial determination of propeller static thrust have a minimal consequence to our final answer.

The actual field assessment reports of performance of these WWII UK Spitfire aircraft can be reviewed at this site.  Calculated performance of two aircraft, K.9793, and P.7280, was compared with field tests.  These two aircraft had published results estimated for zero wind and standard atmosphere given time-at-test conditions at, or very near, sea level.  Calculated results, herein, are taken at sea level. Note: In order to be comparable with field measurements, without error, all results were calculated in English-system units, and are so presented, herein.
For aircraft K.9793, a Mark 1 machine, the calculations were for the aircraft as equipped, and the salient following were ascertained:  Propeller diameter, 10 ft 09 in.  Blade chord at standard 0.7r position, 8.72 in. Rated power at takeoff, 1030 hp @ 3000 rpm.  Reducer, 0.477:1. Static thrust estimated at takeoff, 2413 lb.  Mean thrust thru the take-off run, 2316 lb.  Aircraft take-off weight and pitch, 5,935 lb @ +3.4deg.  Take-off run time at distance, 12.39s at 321 yds.  Documented time and distance, estimated for no wind - standard atmosphere:  12.5s at 320 yds.  Propeller fitted: dual pitch, 3-blade de Havilland with forged and machine-finished duralumin blades.
For P.7280, a Mk-II machine, the standardized aircraft was as follows: Propeller diameter, 10 ft 10 in. Blade chord at standard 0.7r position, 10.5 in.  Rated power at takeoff, 1175 hp @ 2850 rpm.  Reducer, 0.477:1.  Static thrust available at takeoff, 2859 lb. Mean thrust thru the take-off run, 2750 lb.  Aircraft take-off weight and pitch, 6172 lb @ +6.0 deg.  Take-off run time and distance, 9.48s at 215 yds.  Documented time and distance, given for no wind - standard atmosphere:  9.5s at 230 yds.  Propeller fitted: dual pitch, 3-blade Rotol of machine-finished hygdulignum (birch/phenolic-resin laminate) blades.

Summary: So yes, we can calculate realistic numbers for take-off time and distance for the Spitfire.  Several aspects must be considered.  As previously noted, the Spitfire evaluation was accomplished by initial calculation of take-off velocity given the aircraft at its full performance with fitted propeller and Merlin engine.  Initially calculated factors include propeller static thrust and slip-stream velocity for the propeller being evaluated, slip stream influence on lift, and determination of effective lift through the take-off run. Known are aircraft weight and wing area.  We will know propeller thrust and rolling resistance at the start of the take-off run, total aerodynamic drag and velocity at takeoff, and remaining propeller thrust at the end of the run.  We only need to sum thrust and drag and determine take-off velocity to determine distance and time essentially by simple integration. Lift at takeoff is equal to the weight. The slip stream from propeller thrust augments this lift.  Time and distance are derived from the mean of the force of acceleration at run initiation and as the aircraft achieved flight, given the initially determined velocity of the aircraft at the moment of flight.  In setting up the calculation process, we need aircraft pitch to be varied and consequent results noted.  Other than given data regarding the aircraft, we only need, essentially, two resulting numbers in order to determine time and distance to takeoff.  These two resulting numbers must be calculated based upon the developed information.  Those two numbers are velocity of the aircraft at takeoff, and the mean force of acceleration thru the take-off run.

$^*$As a brief trailing end note:  Problem assessment, conceptual evaluation, background research to obtain adequate information upon which to base an analysis, mathematical development and computational procedure assembly, results assessment and validation, as well as development of the text for posting of this answer, required approximately 2.3 months from start to completion.
This was a very interesting question.  Thanks for asking!

Thrust = Power/Velocity

Refers to the energy state of the aircraft.  Units are mv$^2$/s for Power.
People often confuse this "power" with the power output of the engine, which in its most basic form is fuel burned/second multiplied by efficiency factors giving horsepower output,  described as torque at a given RPM.
It doesn't end there.  Now we have to further consider the efficiency of the prop at a given airspeed.  This is why it is very rare to see prop driven aircraft listed as having a given "thrust".
In reality, the horsepower is consumed by the drag of the propeller, or more properly, the drag torque of the propeller, which is drag force summed from a center of resistance not unlike the "sweet spot" of a baseball bat.  Props can also be geared so engine rpm and prop rpm are better matched.

The thrust output is the "lift" the prop airfoil creates.

That is the Force pushing against the inertia, tire friction, and aerodynamic drag of your Spitfire as it accelerates down the runway.
As with the other airfoils on your plane, the wings, quantifying prop "lift" in units of force can be approximated by determining mass flow of air created by the spinning prop as action/reaction forces.  Also, thrust at V 0 can be measured (as with model aircraft) "on the bench" by attaching a scale to the back of the engine/prop mount and applying full throttle.

I propose the following method: the rate of change of momentum has the units of force. When you are standing still on the brakes with full power applied, the prop is taking air which is essentially standing still (zero velocity) and applying a force to its mass which accelerates it to some exit velocity. That force is opposed by the brakes and is equal to the thrust being applied to the airframe by the bearings in the engine case as the prop tries to pull the engine forward.
This means if you know the air mass flow rate through the propeller disc, you should be able to solve for the momentum change and get the thrust that way.
We eagerly await Peter Kaempf's (yes I know it's spelled wrong) opinion on this method!

The spelling is fine for someone on a US keyboard.

what else is going on that makes the true take-off time so much longer than my calculated take-off time?

I think that the poor propeller doing the whole job in front (on in the back) of the piston engine has been totally forgot! The piston engine is producing power but then it's the job of the propeller to convert this power in a thrust. The easiest way to account for propeller thrust, is just by using basic physics.
Being $P$ the power supplied to the propeller by the engine, the propeller changes this power in a thrust $T$ moving the airplane at a speed $V$. In this process, some 20% of the power (at the design point) is lost in inefficiencies which are incorporated in an efficiency factor $\eta$:
$T=\eta \frac{P}{V}$
$\eta$ depends on the speed, with a typical trend visible in the following plot¹:

As visible, $\eta$ depends on the blade pitch (as conventionally measured at 75% of the span) and on the advance ratio $J=\frac{V}{nD}$, where $n$ is the rotating speed and $D$ the diameter. The theory behind this plot and therefore the plot itself is valid for a more or less broad range around the peak: at $V=0$ it gives $\eta=0 \Rightarrow T=0$ which is not realistic since also at zero speed the propeller provides thrust. For that part of the thrust (so called static thrust) you can refer to this answer.
Just try to redo the math taking this into account and let's see if you get a more realistic result.

¹from: McCormick B.W. Aerodynamics, Aeronautics & Flight Mechanics. John Wiley & Sons, Inc.

Prop takeoff calculation is a bit tricky since so much changes between standstill and flight. The most important part is propeller efficiency $\eta_P$ which theoretically would become zero at rest if you use the usual formulas.
The trick is to not take flight speed but the speed in the propeller plane for realistic values. This can be assumed to be half of the eventual speed increase of the air going through the propeller disk $\Delta v$. Then even the efficiency formula for flight can be used throughout the takeoff run. The expression
$$\eta_P=\frac{v_∞}{v_∞+\frac{\Delta v}{2}}$$
becomes
$$\eta_{P_{at\,rest}}=\frac{\frac{\Delta v}{2}}{\Delta v}=\frac{1}{2}$$
which is consistent with reality as the upper theoretical limit for propeller efficiency at rest. Viscous and induced losses reduce this by about 20% for a constant speed propeller and possibly much more for a fixed pitch propeller and the two-position variable pitch prop of early Spitfires.
The proper form of this equation is
$$\eta_P=\frac{v_∞+\frac{\Delta v}{2}}{v_∞+\Delta v}$$
Strictly, this is the correct efficiency equation for the full speed range, but was a bit impractical in the time before computers because using it requires an iterative process in which thrust and $\Delta v$ must be balanced. Therefore, all books of authors who blindly copy what has been published before will show you only the first equation. The error is small once $v_∞$ becomes large, especially since $\Delta v$ drops with increasing flight speed.
But there are more losses. Once the prop is running, the airplane will produce aerodynamic drag even before it has started to move. The propwash flowing over hull, tail and wing root will produce viscous drag which of course only grows once the airplane picks up speed.
A very interesting paper on the efficiency of an installed propeller can be found here. By comparing the gliding qualities of a Luscombe 8E with drag in level flight, the authors measured the drag increase of the airframe caused by the propeller wake. A drag increase of 30% from the propeller effect was measured, resulting in an overall propeller efficiency of only 62% when the propeller alone had an efficiency of 81%.
The smallest bit that is still missing is rolling resistance. This adds maybe 2 - 3% of overall drag and goes away as the wing lifts the airplane up. The price for this is induced drag on the wings, of course.
Since you post on Stack Exchange, it is safe to assume that you have a computer at hand and know how to use it. Write a small program which calculates all forces and integrates them for a small time step, then repeat at the new speed until you reach lift-off speed. The correct take-off time and distance will be a byproduct of this procedure.

Please see attached article graphs.

@RobertDiGiovanni The author also makes the mistake of zero prop efficiency at zero speed. This contradicts his own thrust curve which looks quite realistic.


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What is the speed of anti-ice bleed airflow in helicopter turboshaft engines?
If we consider the most modern turboshaft engines, what will the approximate speed for the bleed airflow when it's routed from the latter stages of the engine and supplied to the areas that require anti-icing. Apologies, if the question is a bit broad.

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer.

Don't apologize for breaking the rules, [edit] your question to follow them.

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

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Cessna circling Penrith regularly at night
For the past year we have had a Cessna circle our local area (Penrith NSW), starting early in the night and often going well into the early hours of the morning. This will happen once a week to fortnight.
On flight tracker, it shows as a Cessna C208.
As a prop plane, it's extremely loud and drones on. My fiancee cannot sleep from the sound.
What could be the cause of this plane flying so loudly in circles, and so often? Finally, how is this legal?
Thanks!

Usually these questions are simple things like flight training, aerial photography, etc. But looking at the track it's just endless circles at 5000ft directly over the town. Perhaps a very spiteful divorcee? Try and figure out where they land and call up the airport...

It is most likely an aircraft from NSW Police. I say this because the registration is blocked on FlightRadar24, knowing that they have a fleet of Cessna 208 aircraft based out of Bankstown airport, and because at that area at 5000ft is in what is known as controlled airspace, meaning nobody can fly there without air traffic control permission, which is hard to get when Sydney international airport is open (admittedly it is under curfew from 11pm - 6am with some exceptions).
One of the activities listed on their website is Forward Looking Infra Red scanning, presumably looking for some bad people.
Unfortunately, being a police aircraft, you're unlikely to get it stopped. Consider perhaps contacting your local NSW parliament representative?

Reminds me of the scene in Captain America: Winter Soldier when infrared scanning satellites and flying guns pretty much presume everyone is a bad guy.

Thanks Ben! I think you're right, pol air 7 being most likely. This aircraft was billed as being perfect for "remote areas" due to its long fuel range... Unfortunately, police are a law unto themselves so I think your suggestion to write to the local member is the only non violent way to address this. I note a local man was shining a laser at this exact aircraft earlier in the year (and had riot squad bearing down on him very quickly)

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How does an aileron actuator work considering the following?
My question is how does an aileron work considering it rotates around a fixed point and the actuator arm moves in a straight line. If what is shown in the first picture I have provided is true, then how can the actuator arm move the lever that moves the aileron considering the fact that the lever moves in a circular direction and the actuator arm only moves in a linear direction? If the picture is right doesn't that mean the lever that moves the aileron would have to change length to be able to move in a circular direction since the actuator arm moves in a straight linear direction? Please let me know if you need clarification.

Does this help:  https://aviation.stackexchange.com/questions/78967/how-do-linear-actuators-for-flight-controls-work-in-fighter-jets?rq=1

Maybe if you see it it's easier (it's a spoiler actuator but it works just in the same way)

There are many, many places where machines have to convert "linear" motion to circular motion. The connecting rods of a steam locomotive or an automobile engine are examples of the "dogbone link" mentioned in an answer below. The rear shock absorbers of a typical motorcycle work like the actuator show in your figure.

You will see two methods of providing articulation for the hydraulic cylinder to accommodate the vertical shift of the attachment at the aileron as the aileron moves.
In one method, one end of the actuator has a pivoting attachment to the structure and the other end a pivoting attachment to the control surface. The entire actuator moves in a small arc defined by the relative up and down shift of the attachment at the aileron as the aileron moves.
An alternate method is used if you need the actuator to be rigidly attached to the structure (usually because of space considerations that make it undesirable to have an actuator that can articulate up and down itself, or the design of the input linkage requires the actuator to be stationary).
The rigidly attached actuator is connected to the control surface by an articulating link, sometimes called a "Dogbone Link". This actuator operates on a fixed axis, and all of the geometric compliance is accommodated by the two pivot bearings in the dogbone itself.

The actuator extends linearly indeed, and is mounted on both sides with bushings that allow for rotation. As can be seen in above picture, from this site.
So when the actuator extends:

the end at the actuator arm follows a circular curve;
the other end stays fixed in one point;
both ends rotate around their bushings.

What about the actuator on the top, the 2 bushings aren't parallel. Was this used for some other flight actuator?

@JohnSmith That aileron actuator is not clamped in on the left hand side, it is mounted such that rotations about the left hand side attachment point are allowed.

But it couldn't be used for an aileron because the right hand side of the top actuator needs to rotate in a parallel direction to the left bushing, right?

@JohnSmith I think what Koyovis is saying is that the bushing on the left can rotate so that it is parallel to the one on the right. (I'm not completely sure I'm understanding correctly, though.)


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Failed the private written test and I need an authorized instructor's statement, is there any way I can get that online?
I failed my Private written (I was being stupid and overconfident) but I got my endorsement online, I need an AUTHORIZED INSTRUCTOR'S STATEMENT to retake it and my original instructor is unavailable. Is there a way to get that online or do I have to find a new instructor to go over and sign off on it.

Hi ZKelly, welcome to Av.SE.  It looks like you have two separate accounts; please see https://meta.stackexchange.com/help/merging-accounts for steps to merge them.  Also, please be aware that "resource location" (i.e. where can I obtain ___) is off-topic here, so you probably won't be able to get a list of online instruction providers who do that sort of follow-up (if such exist -- I'd be a little doubtful).  Although I think there are enough other aspects to your question that it's okay as it stands.


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How can I see the track of a flight from a year ago?
I want to get tracking data from a particular GA flight that took place over a year ago. I know the departure airport and the city the plane landed near, and I can narrow it down to a few possible arrival airports. I know the day the flight landed, and it took off either early that morning or late the previous night. I also know the airplane's registration number.
How can I get more information about this flight?

Have you tried Flightaware or other such apps?

I don’t know about ADSBexchange, but the main commercial sites/apps (FlightWare and FlightRadar24) only provide a few days of history for free.  Then you pay for increasingly expensive subscriptions in order to get longer histories.  But I believe they both max out at one year.


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How much admin time required for owning a light aircraft?
If I were to buy a light aircraft, I imagine there would be quite a few administrative things I'd need to keep on top of. Maintenance, insurance, logbooks, accounting, etc.
Is there anything I haven't thought of? For example, would I also have to inform my life insurance provider or possibly find a new one?
I'm basically wondering how much time I would have to put into owning an aircraft above and beyond the time I spend flying it.

Hi Kevin, welcome to Av.SE! Nice first question - I expect we'll have several folks here who can provide examples of what they've experienced with their aircraft.

I'm guessing by the term "light aircraft" you are looking for costs in a country other than USA? what country, and what plane?

Hi @tedder42, it's actually the USA I'm interested in but now you've got me asking a new question... what do you call light aircraft in the USA? :) I was thinking of a 4 seater, something equivalent to a C-172. I've found a few resources that describe costs in terms of money, but the thing I'm wondering about here is the cost in terms of time spent doing administrative tasks.

Not that much.  Ownership is more of a money eating machine than a time consumer for the owner who has someone do their maintenance.
Your AI/A&P takes care of your tech logs. You just fly it, schedule maintenance with them, deal with your insurance.  Your LI may or may not have a flying exclusion. Examine the contract. If nothing is there, you're good to go. If it is, you may have to pay extra.
I do my own maintenance and annuals on my homebuilt so there's that work once a year and from time to time for unscheduled fixing and tinkering, but otherwise, not much direct activity.
A typical owner just flies and pays bills. Lots and lots of bills. If the financial burden is light relatively speaking, that is, it's money you could use to light a cigar without giving you a heart attack, you can enjoy it.
If the costs are a burden and make you sacrifice other things that you'd rather not, it takes out most of the enjoyment.  The old saying goes, like with large boats that are bit more than someone can afford: the best days are the day you bought it, and the day you sold it.

I agree with this. A couple points: If you keep a good log as you go - flying hours, database updates, squawks (to pass to maintenance), it makes life easier. I found time-wise, updating the databases every 28 days to be a nuisance. The time is highly variable depending on your avionics, but it's generally less than an hour. If you can assist your A&P on the annual inspection, it'll cost you a day or so of your time but it'll save your A&P some time and you some $ and you'll learn a lot. A lot of the inspection involves removing and re-installing inspection panels.

In Canada the owner has to maintain a Journey Log by regulation recording all flights, to, from, times, air time, flight time, total airframe hours, snags/rectification, etc. When you buy an airplane in Canada you have its operating record for its entire life right in front of you.

My life insurance specifically excludes flying other than a passenger on a commercial flight, so I always try to ensure I don't die in my spamcan.  It's worked so far...


text:
How many commercial or ATP pilots have lost their medical yearly from 2015 to 2022?
I wold like to know how many pilots have lost their medical that prohibits flying on a commercial or ATP, year to year from 2015 to 2022.

I doubt this is public information...

You need to write the FAA to see if that statistical data is publicly available.


text:
Why does a paraglider not need a vertical stabilizer
I am looking at a question I found in the internet: why can a paraglider fly without a tail rudder and elevator assembly whereas a normal fixed wing aircraft can't?
One part of the answer is , the elevator helps to control pitch whereas the rudder controls yaw for a fixed wing aircraft.
A paraglider has flexible wings that can be deformed to control both the axis.
The other part that is vexing me : the tailplane also provides stability, not just axis control. So why is such (lack of) stability providing equipment not detrimental to a paraglider?
Even birds and there flexible-ish wings need a tail.
How do I even approach this question? Is it because the bird wings are not that flexible, and so the large tail feathers are necessary for fine control?
What about the fixed wing aircraft? Is it because the tubular shape causes the centre of lift and centre of gravity to be far away and so a tailplane is mandatory?
I would like a full mathematical answer please- thank you?

What you are looking for is termed "longitudinal stability analysis". Here you can find a very good overview.

Because of the "pendulum effect".  The center of mass is way below the wing.

@quietflyer  True, and the pendulum effect is enhanced by a HUGE anhedral wing.

An aerofoil is unstable in airflow; if it pitches down slightly then the centre of lift will move rearwards, amplifying the downward pitch.  In early hang gliders this could result in a ‘luffing dive’ which was typically unrecoverable.  This was resolved by adding luff lines that pull the sail into a reflex form.  In paragliders, pitch stability is provided by placing the pilot far below the wing.  This is somewhat contentious because ‘pendulum stability’ is typically negligible in rigid aircraft where the c of g is not far from the centre of lift.  Yaw stability is inherent because of the large side area presented by the curved wing.


text:
Air accidents that was caused by experienced pilot's wrong assumption that the controls were identical in different aircrafts
I am compiling an essay about pilots making wrong assumptions about controls and layouts and behaviors of aircrafts.
The cases I am looking for should demonstrate that highly experienced pilots in one (sub)model of a specific aircraft assumes that another (sub) model or similar aircraft would be pretty much the same and/or their training/muscle memory/experience kicks in - and they activate the wrong thing leading to a disaster.
One example that i could find is : Lokomotiv Yaroslavl plane crash
One of the speculative reasons is/would be that the brake pedals of the Yak42 was not designed for the pilot's heels to rest - in contrast to a very similar Yak40. That lead to insufficient takeoff acceleration, runway overrun and disaster.
Another one that could fit the bill is the Überlingen disaster, where their experience led to the decision of listening to the controller and not the TCAS - which made the worst disaster over post war Germany - but I would like to focus on cases, where the experience was detrimental with respect to the aircraft itself.
Are there other similar cases, where the pilots experience (but not "tailored" experience) led to immense fatalities?
Thank you.

Might look into F-104 crashes... some had upward firing ejection seats, others had downward firing.  The latter had to roll inverted if ejecting at low altitude, which would be a really awful thing to do in the former.

As you rightly said, I don't think Überlingen fits the bill. The cause was incompatible expectations between the two crews, caused by missing international standardization and guidance. If anything, the cause was cultural, with the "Western" pilots being used to trust the machine and the Russian ones being indoctrinated to blindly obey authority always, no questions asked. And, of course, the whole tragic mess at the ATC site.

Hasn't lead to disaster as far as I know but the EXPED button on the A320 MCP was replaced by the ALT button on the A330/A340/A380, which led to confusion for cross-fleet flying, some airlines banned its use, and eventually EXPED was deleted.

Aeroflot Flight 821
The attitude indicators used on most Russian aircraft have a different functional configuration from those used on Western (e.g., Boeing, Airbus) aircraft. This was a major cause in this accident as the pilots previously flew Tupolev and Antonov aircraft and the accident was in a Boeing 737. While this was not the sole cause of the accident, it contributed to loss of spatial orientation which was a major cause of the accident.

Have a look at the Kegworth Air Disaster  https://en.wikipedia.org/wiki/Kegworth_air_disaster where the pilot shut down the wrong engine because of differences between the 737-300 and 737-400.


text:
Is there a historical source of all flights from a given airport?
I am trying to find delay data on all flights that departed from Malaga (AGP) airport on 29th September 2022. I have tried a trial of the highest tier subscription for Flight Radar 24. However, this is great for single route history, or short-term history for an airport (even on their highest tier, I could only go back a day or two for a whole airport), but I could not find a way to obtain the information I'm after.
I have looked at other sources such as Flightstats, Flightaware etc. (the first few on Google), but it is not clear whether they would provide this information. Does anyone have a suggestion on where I might find this? Ideally, the source would have a trial if the data is paid-for, as this is a one-time request only. I would consider paying if that's the only option.

There's a couple of options here.
You could try the most direct route in contacting the airport directly (they have an email address and phone number) and ask.  Since this is only one day of data, you might get lucky (depending on how friendly they are) and get exactly what you're looking for.
https://www.airport-malaga.com/
The other option is a provider.  OAG offers historical data and have a trial available that might give you what you need.  The application form for this requests a company name, so this might not be applicable for individuals.
https://www.oag.com/historical-flight-data
As you've doubtless seen, historical data from FlightAware is expensive.  Obviously, historical data has commercial value in terms of analysing travel trends.


text:
Does a 120cc engine burn 120cc of fuel a minute?
I have heard that a 120cc engine burns 120cc or 120ml of fuel per minute. Is this true?

No, it is not true. Where did you hear this?

120cc is not fuel burn rate. It is the total internal size of your engine's cylinders. When running the engine does not fully flood the cylinders with fuel. Most of what the engine breathes in is air, not fuel. This is true of all internal combustion engines be it for aircraft or motorcycle or car or boat or lawnmower.

@slebetman Actually, it is the difference of the spaces in the cylinders between the downstroke and the upstroke position. In other words, it is the space swept by the pistons.

@PeterKämpf also, its the sum of all the cylinders, so a 2 cylinder 120cc engine has 60cc per cylinder, whereas a single-piston 120cc engine has a 120cc space in the one chamber

@Criggie No. Please read how it is defined, for example here: https://en.wikipedia.org/wiki/Engine_displacement

@PeterKämpf "Displacement = stroke length x pi x bore^2  x number of cylinders"  not sure what we're disagreeing on here.

@Criggie Only when the combustion chamber at the top dead center is zero is displacement equal to the "sum of all cylinders". I don't think that makes a viable engine.

Bore^2 is wrong.  It should be (1/2*bore)^2.  Bore is the diameter of the cylinder, and the swept area is proportional to the radius^2, just like the area of a circle.

That's like a push mower engine, nothing that would be on an aircraft.

@PeterKämpf: I think you may be misunderstanding what Criggie was saying? As I read his comments he's just emphasising the "x number of cylinders" part of the formula; there's no contradiction between what he's said and what you're saying, as far as I can see.

Ha! I remember when I was a kid my dad would complain about how wasteful 4.4L V8 (car) engines were and in my head was imagining over a gallon of fuel being burned at a time...

I would say it's unlikely, and is certainly not a rule of thumb. Fuel consumption depends on many variables, including:

Engine fuel and cycle type (e.g. 2-stroke petrol/gasoline)
Engine speed (RPM) and load
Altitude (air density, temperature, pressure)
Fuel/air mixture (rich/lean/stoichiometric)

To be clear, “120 cc” refers to the engine displacement, i.e. the combined usable volume of the cylinders (minus the combustion chamber volumes)—basically the amount of air the engine can “breathe” in one cycle. 120 cc would be a pretty small engine, so let assume it's a 2-stroke (1 cycle per revolution) cruising at 2500 RPM at a few thousand feet. Here I'll use the Frink engineering calculator syntax. Air density would be about 0.9 x that at sea level, so the mass airflow could be calculated as follows:
engine_displacement = 120 cc
engine_speed = 2500/min
cycles_per_rev = 1
air_density = 1.204 kg/m^3 * 0.9

engine_airflow = engine_speed / cycles_per_rev * engine_displacement
mass_airflow = engine_airflow * air_density

mass_airflow -> g/s
5.418

For the fuel consumption, let's assume a stoichiometric air-fuel ratio:
fuel_flow = mass_airflow / 14.7
fuel_flow -> g/s
0.36857142857142857143

Converting to volume flow rate:
gasoline_density = 750 kg/m^3
fuel_flow / gasoline_density -> cc/min
29.485714285714285714

So around 30 cc/min for those particular conditions. You could imagine considerably more for a high-power climb at very high engine speeds (potentially even more than 120 cc/min).
Thanks for the interesting question!

Nice answer - welcome to Av.SE!

How do you convert mass airflow to G/s

It's not a direct conversion, as it depends on the air-fuel ratio and the density of the fuel. If you know those, it's mass airflow / air-fuel ratio / fuel density.
(Note that the air-fuel ratio is a mass ratio, as that reflects the amount of matter in the chemical reaction of combustion, and isn't affected by temperature, pressure, density, etc.)

This is possible, but still quite extreme fuel consumption.
120ml == 120cc gasoline are ~90g by mass
90g gasoline need about 1320g air in order to fully combust
1320g air are about 1.08 cubic meters at sea level and 15C
Assuming volumetric efficiency of 1.00 (possible with a supercharged engine):
1.08 cubic meters need 9000 cycles of 120cc displacement.
This amounts to 9000RPM for a 2-stroke engine or 18000rpm for a 4-stroke engine.
Rather at the high end, but not unseen in a high-performance engines (e.g. sports motorcycles).
If you allow for a rather rich mixture (e.g. AFR of ~11.0, one will possibly need to run the engine rich at these performance levels) the engine may even not be supercharged and will still be below 10000/20000rpm.
But since we are on the aviation.SE I state that I don't want to fly with an engine pushed this much to the technology limits.

No, a 120cc engine does not burn 120cc of fuel per minute. It means that the  sum of the volumes of space swept by the pistons in each of the cylinders equals 120cc. This is called engine displacement.
The amount of fuel burned will vary with engine power output (i.e. throttle) and likely other factors. An engine with a higher displacement will likely burn more fuel than a smaller one, but if other components or the circumstances are different, this is not necessarily the case.

It is not the sum of all interiors, but the sum of the space swept by the pistons. Please compare to the Wikipedia article your linked to.

@PeterKämpf I fixed it

The fuel burn of a very small 2 stroke with a crude carburetor is roughly .6 lbs/hp/hr. You can work it out from that.  A 10hp engine running at a cruise power of 70% will use .6 x 7 = 4.2 lbs/hr at cruise power. Which converts to 44.12 cc per minute, or 2.65 L/hr.

Do you have a source for these numbers? Is this a lawnmower or outboard?

Not at hand. I read it many years ago. It basically goes: 2-strokes in general 5.5 to 6.5, carbureted air cooled 4 strokes around 4.5 +/-, injected 4 strokes around 4+, liquid cooled injected 4 strokes like car engines high 3s, diesels mid to low 3s. Turbo props are similar to 2 strokes around 5 to 6.5.  The aircooled Lyc O-290 in my plane is around 4.4 from the performance charts. The turbo compound Wright R3350s were in the mid 3s, almost like diesels.  2 strokes are about the least efficient int comb engines there are. They are used for the power to weight.

@DDuck see here https://en.wikipedia.org/wiki/Brake-specific_fuel_consumption with pretty good table of assorted engines

No. "120cc" is the swept volume of the cylinder
It represents how much volume of air the engine can breathe per cycle, not how much fuel it might use.  It says nothing about fuel consumption.
The cylinders are round and go up and down.  They're taking cylinder travel x the surface area of the circle

pi x radius2 x travel.

As a thought exercise, 120cc is a small motorcycle engine, way too small for cars. If it was using 120CC/minute say at 60 MPH/100 KPH, that means 1 litre in 8 minutes/miles = 8 liters per 100km.  Or 32 MPG. Awfully low for a 120cc bike. There are large pickup trucks that get better MPG/km/l than that.
However that fuel economy might make sense on an airplane. A very small airplane.


text:
What is the glide ratio of the Ozone Mojo Power 2?
The ozone mojo power 2 is a paramotor wing aimed towards beginners. I am interested in purchasing it, however I have not found any data on the lift to glide ratio. Does anybody know this info?

I believe EN-A level wings are generally about 7-8.  EN-Bs around 9, EN-Cs/Ds around 10+.  I think you'll find they are all so closely matched performance wise, handling qualities become more important. As a power/glider pilot who wants to protect my legs and back and having watched enough vids of the results of collapses close to the ground,, passive safety would be priority #1. I'd like to get into paragliding, but I have a plantar fasciitis problem with my left foot which prevents me from running.

Have you tried paramotor trikes? It severely reduces running  and stress on your legs. Hope you get better.

Thks. The trike config takes away some of the bird like feeling of just being in a harness. Might as well move up to a hang glider trike or an ultralight.


text:
Would the dle 55ra engine be enough to power my paramotor(ducted fan)?
I plan on using the dle 55ra engine to produce 5hp at 8,500 rpm. I am using it in a paramotor, and to save space I am considering using a ducted propeller similar to the Fulcrum power pod. The duct intake would have to be directly behind my back, so I plan to add around 6 inches of space to allow air intake. The engine would be mounted on my back facing into the intake. Effectively, it would be a normal paramotor with a duct around the fan blades. My estimated duct wall clearance is 1 or less mm. I need to produce 15kg of thrust, so how would I design this fan? I considered using a heating duct, which is an 18-16 reducer. This would allow me to get some benefits of the thrust tube and would make it easy to have a front-mounted propeller. Would this propeller work? It is 18x7"
TLDR: Would my combination of propeller, duct and engine produce 15kg of thrust.
Thank you!

Paramotors require at minimum 125lbs of thrust to have any kind of performance for climb. Most are around 140-175 lbs. 35lbs won't even get you off the ground.

@GregHewgill  It's a paramotor question. Paramotors are airplanes.

@John K, I don’t think I mentioned this in the question, however in a separate post other users helped me calculate that 5hp was enough to get 82kg full weight in the air, with some extra leftover.p(~2.5 extra hp). If my wing has a glide ratio of 7:1, then 82/7= 11.7143kg required for level flight. The approximately 3kg of thrust left would be for climbing. Would I need more? 20kg of thrust in total would probably be good.

+Lelotz I already factored in climb ratio, as me and my  setup weighs 64kg now, and I am accounting for future weight gain. 64/7=9.1429, so 15 is more than enough

You will need WAAAAAY more.  This harness/power pod appears to be exactly what you are shooting for:   https://www.fulcrum-engines.com/.  Its engine is 25HP to make 110 lbs thrust, about the absolute minimum.  110lbs is not much for 25 hp, ducted fans being somewhat inefficient.  Regular PMs make 150-170 lbs with the same engine using a large prop and reduction drive. That's the price paid for the ducted fan.

@JohnK: My apologies, I did not read the question carefully enough.

@John K. Why would I need so much thrust? I do not weigh that much and at min 25kg is probably the best solution. I average a bunch of paramotors together, their thrust/max weight and got an approximate value of 0.35. 25/82 is 0.3 and they are not far apart. 170 lbs is enough to make me hover and climb quickly

Well, about the smallest paramotor you can get is the Atom 80, which makes 120lbs thr with a large prop and belt drive. That's about the bottom end, meant for girls and very light males.  You need the thrust to be able to escape from bad situations close to the ground when your airplane is so hypersensitive to air movements close to the surface.


text:
PROB30 in a TAF as an alternate still legal to use
Does a PROB30 in a TAF report affect your ability to use an airport as an alternate if conditions still remain above landing minima but less than alternate minima.
Example:
TAF XYZA 061335Z 0615/0715 33012KT 9999 SCT015 BKN025 TX02/0618Z TNM02/0715Z
TEMPO PROB30 0615/0703 800 -RADZ BR BKN004 OVC013
BECMG 0703/0705 03008KT
BECMG 0709/0711 CAVOK
Operating minima at Airport XYZA is 550M/200'.
Flight ETA falls within 1500/06-1500/07. Can you still use the airport as an alternate airport?

Yes. Exactly. Canadian AIM states if the PROB30/PROB40 is below alternate minima but not landing minima for the airport, you can still use the airport as an alternate. My understanding of NWS is that it doesn't use PROB40, so is this legal in the US? UK/ICELAND/GREENLAND/CANADA(verified).

I guess this source is not reliable, then:  https://training.weather.gov/nwstc/aviation/TAF/lesson4/prob30.html


text:
How can I calculate the thrust of a propeller?
I am working on designing a paramotor, and need to calculate estimated thrust for the propeller. Is there a calculator or equation where I can input a diameter, pitch and rpm and get a ballpark thrust estimate and power consumption. I would prefer the thrust estimate to be lower rather than higher

If you can't find the formula, paramotors with decent size props driven through a reduction drive are about 5-6 lbs thrust/hp.  You could use 5 lbs and be sure of being conservative.

Peter Kaempf on this site will surely know the formula.

It’s a little bit hard to understand. Can you help me summarize?

The relation between thrust $T$ generated by a propeller and the power $P$ which has to be supplied to the propeller, is given by two distinct equations, one valid at zero flying speed and one valid at the "design speed". Let's see these two equations.

Zero speed
For this part, I gave already here exactly the needed equation:
$T=\eta (Pd)^⅔\sqrt[3]{½\rho π}$
where $\eta$ is an efficiency factor of some 0.6 to 0.8, $d$ is the propeller's diameter and $\rho$ air density.

Design speed
The propeller simply changes the engine's power $P$ in thrust $T$ pushing the airplane at a speed $V$. In this process, some power is lost in inefficiencies which are incorporated in an efficiency factor $\eta$ (different from the previous one):
$T=\eta \frac{P}{V}$
$\eta$ depends on the blade pitch as measured at ¾ of the span and on the advance ratio $J=\frac{V}{nd}$, where $n$ is the rotating speed [rev/s]. The precise trend of $\eta$ is supplied by the propeller's manufacturer but it has a typical shape like the following one¹:

Note that this plot becomes meaningless for $J$ smaller than some 0.2.

Let's do the math using for example the following values:

$\eta$@0speed, 0.6 (to be conservative); if there's a gearbox between the engine and the propeller, some power is lost there, let's say some 20%; we incorporate this loss in $\eta$ that therefore becomes $0.6 \times 0.8=0.48$.
$\rho$@sea level, 1kg/m³
Engine's power 3kW@2000rpm i.e. 33rps.
Propeller's diameter 1m and 40° twist@¾ span.
Cruise speed 60m/s.
$J=\frac{60}{33}=1.8$; from the plot we get a relevant $\eta=0.8$.

So:

at 0speed thrust is $0.48\times3'000^⅔\sqrt[3]{½π}=116N$
at 60m/s it is $0.8\frac{3'000}{60}=40N$

Between those two speeds thrust can be simply linearly interpolated.
Obviously these are all theoretical values which have to be tested in reality. Anyway they can be used to ballpark estimate the main geometry of the propeller and the needed power.

¹McCormick B.W. Aerodynamics, Aeronautics & Flight Mechanics. John Wiley & Sons, Inc.


text:
How much horsepower do I need for my paramotor?
Assuming my full kit and I weigh 82kg total, I have estimated that I need 5hp. To calculate this, I used the formula:
weight force (W) is F = m g
W= 82kg*9.8 ms/2 =401.8 Newtons
Now, if the drag ratio is 9:1 that means that there is also 401.8 Newtons of Lift (L) and 45 Newtons of Drag (D): D= L / (L:D) 45= 401.8/9 So if 45 Newtons are required to be overcome drag, this is the thrust the engine must produce. Using a rule of thumb of 11 Newtons per Hp that equals 4.05 Hp. I rounded this up to 5hp for climbing.  Is my math correct? 5hp  seems low compared to paramotors on the market with 20+hp. I would also need approximately 10kg of thrust, 82/9=9.1 rounded up to 10

An easy way to calculate the power is to think in the 'gravitational power' of the aircraft in a glide. If the horizontal speed is S, and the glide ratio (L/D) is R, the vertical sink speed V = S/R. Now, the power P involved is the weight W of the aircraft times that sink speed => P = W * V. For a given mass and L/D, the 'gravitational' power involved in the glide is the same s the power you'd need for straight and level, unaccelerated flight...

Make sure you have excess power to climb.  You can figure ROC as a mirror image of glide ratio.  So if you climb 1 to 9 @ 40 knots, climb vertical rate is around 4.5 knots or around 450 feet per minute.  To provide safety margin,  double hp required for level flight.

With a ratio of 9:1 (rather high for a paramotor) and a mass of 82 kg, you have a 'gravitational power' (in a glide at 60 km/h = 16,7 m/s) of 82 * 9,8 * 16,7/9 = 1491 W = 2 HP. That, also for prop-driven s/l flight at that speed and L/D. Thrust would be 1491/16,7 = 89 newton.
But if you use an engine with a prop, you have to take into account the propeller losses. If the prop efficiency is 75%, you would need an engine power of 2/0,75 = 2,7 HP.
Probably, the L/D is not 9, but much less, so you might need more power...

I plan on flying the Ozone Mojo Power 2 wing, I couldn’t find any data on the glide ratio. A quick google search told me that an average is around 9, but would 7 be better? I have heard it’s always better to be over than underpowered. Would a 3 blade  ducted fan of 16x10” with a duct wall clearance of 1mm be enough? I plan on using the dle 55RA engine, as it can put out 5hp at 8500rpm

@Lelotz Let me suggest not to use the comments to make questions. Instead, you should put them in the Aviation SE...

If the average speed of the wing is 40km/h would I change the 16.7 to a 11.1?

@Lelotz 1 m/s = 3,6 km/h


text:
What is the real formula for Pressure altitude?
OK. The formula we commonly know is (29.92-Altimeter)x1000+Elevation.
However, the E6B calculator I have does not fit this formula at all. For example, if the airport elevation is 175 ft and the barometric pressure is 30.21 in.hg, the general formula gives -115 ft. But on the calculator it comes out as -91ft.
I know that the calculator is obviously more accurate. And recently, I heard that the formula that is commonly known came out for convenience. So, how do I get the real formula?

Maybe this answer can help.

Check this: https://aviation.stackexchange.com/questions/71962/what-formula-to-use-for-calculating-pressure-altitude#91615

The approximation is used because the actual formula is too complicated to commit to memory, and too complicated to do by hand in flight.  There’s an exponential aspect to the atmospheric pressure lapse rate that makes it a hard problem to solve on paper on your knee board.  But most of our flying is done in conditions where pressure altitude will tend to be in a range of -1,000 to +1,500 feet of our indicated altitude, and we’re using pressure altitude to look up performance data that is likely listed in 1,000 or 2,0000 foot increments in the tables.  So a rule of thumb that is accurate to within 100 feet was always good enough.  But when you’re doing other calculations and you want to start with a more precise pressure altitude, then:
Pressure Altitude = Indicated Altitude + $145442.2*\left(1- \left( \frac{Alt Setting}{29.92126} \right)^.190261\right)$
AltSetting is the altimeter setting in the Kollsman in inches hg.   Indicated Altitude and resulting Pressure Altitude are feet.
It’s non-linear:

Altimeter setting of 30.92126 = -912.6 feet of correction.
Altimeter setting of 28.92126 = +937.6 feet of correction.
Altimeter setting of 27.92126 = +1,925.2 feet of correction.

In your example, 30.21 works out to be -265.99 feet, plus your elevation of 175 = -90.99 feet, almost exactly what you got.
If I wanted something more accurate than the old rule of thumb, but still as easy, I think I’d use (29.92-altimeter) x 940 or something like that.  That’s accurate to within 40 feet or better across almost the entire normal range of altimeter settings.

I'm sorry for late response. Please understand that there is a time difference. you are so kind. I put in the formula you gave, I get the correct value. My questions have been resolved. thank you.

@rockybear There’s no need to apologize, this isn’t a real-time platform and there’s no expectation anyone is waiting by their computer for an answer.

@MaxR, thanks for the great answer. This does seem to match the table the FAA provides in their testing supplement. Do you have a reference for the original source?


text:
Why do they say PBN is under RNAV?
PBN exists under the umbrella of area navigation (RNAV).
[FAA AIM, Section 2. Performance-Based Navigation (PBN) and Area Navigation (RNAV)]
As far as I know, RNAV and RNP exist under PBN, and as an abbreviation of Performance based navigation, I know that they belong to a large category. However, FAA AIM states that PBN is under the umbrella of RNAV. Any detailed reasons?

@DeltaLima, Why did you delete the answer as rude or spam?  It wasn't correct in my opinion, but definitely not rude.

@MichaelHall: Because it linked to the registration page a government program, so spam.

@mins, I cannot see what you are talking any more about to verify, but it asks a question about the FAA, a government agency/program, so a link to an FAA page (if that's what it was) isn't spam by the definition of the term.

@MichaelHall: No that was the government of Pakistan and not related to aviation at all. The link was hidden behind "Need More detail check here", obviously intentional.

Oh, ok.  Nevermind then...  Thanks.


text:
What is the difference between IMU and IRU?
During my academic studies, I am studying the INS part. First of all, INS means a comprehensive category combined with IRU and inertial navigation computer. The IRU is a sensor that uses a gyroscope and accelerometer to detect rotational attitude changes.
But the problem is the IMU. It is written that the IMU also works by detecting changes in the attitude of the aircraft, including the gyroscope, accelerometer, and even the magnetometer.
So, are IRU and IMU the same concept? I looked elsewhere, but I couldn't figure it out.

It looks like IMU is a generic name and they both are interchangeable.
From Wikipedia: An IMU (inertial measurement unit) is an electronic device that measures and reports a body's specific force, angular rate, and sometimes the orientation of the body, using a combination of accelerometers, gyroscopes, and sometimes magnetometers.
IMUs are often incorporated into Inertial Navigation Systems which utilize the raw IMU measurements to calculate attitude, angular rates, linear velocity and position relative to a global reference frame.
Source: https://en.m.wikipedia.org/wiki/Inertial_measurement_unit
Here says: Another name often used interchangeably with IRU is Inertial Measurement Unit.


text:
How often do weather balloons drift outside country borders?
The Chinese balloon incident made me wonder how many high altitude balloons have drifted outside their own country border. It seems weird that a lot of weather balloons suddenly drifted into other countries unannounced, when GPS is cheap and light. It seems that the search engines are still flooded with news of the incident and the only other incident I can find is the 1998 Canadian weather balloon which rode on that wave.
I am quite curious on how often do weather balloons get lost so badly it drifted outside their own country, or maybe even balloons that failed to pop for so long. It can be simply crossing the land border of a neighboring state.

Re "I am quite curious on how often do weather balloons get lost so badly it drifted outside their own country, " -- actually, if the country is small (like in Europe), they are not supposed to stay in the country from which they are launched.

@quietflyer People in the EU can move quite mobile, but for closed border countries how do they retrieve the balloons, do they have some agreement with each other?

@R3G3N3R4T0R For traditional weather balloons, they're considered disposable. The US National Weather Service says they get less than 20% back, and if it ends up in Canada or Mexico, to throw it out. There's no expectation they'll recover radiosondes launched from coastal sites that drift over the sea or launched from places like Hawaii and Alaska. This is distinct from scientific high-altitude balloons, which are much larger.

I know the Royal Dutch Meteorological Institute (KNMI) in the Netherlands attaches information to their balloon radio sondes in German, French and English.
With balloons frequently travelling >200km, and the country's border being < 200km away from the main meteorological centre in any direction, the balloons crossing into a neighbouring country is a pretty normal event.


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Why are catapult carriers' decks flat and leveled?
I know the Russians seems to be planning some sort of catapult then ski-jump system, but I am not talking about that.
In the history of all aircraft catapult, they all seems to be launching the aircraft leveled instead of pitching upward. From old seaplane launcher to modern supercarriers. Is there a reason why no one is doing it? Does it make it cope harder in rough seas? Does it make it more complex, procedure and mechanism wise? All I can find is someone on Quora saying that they work better leveled without any explanation or citation.

What benefits do you think there might be from an angled deck?

I think it can give the plane a bit more angle of attack. From what I have seen there seems to be some kind of procedure that requires you to quickly rotate the aircraft at the end of the launch. I also think it may allow launching heavier aircrafts as the upward momentum should give a little leeway before reaching take off speed.

You get greater angle of attack, but at the expense of speed and mass, which means it is worse for heavier aircraft.

are you asking about the entire flight deck or just the catapult area?

I can only imagine the engineering complexity to have the catapult track bend to go up the ski jump. Additionally, you'd have the nose gear of the aircraft compressing as it hits the bottom of the ramp instead of staying fully extended as it's pulled. Also, since all carrier aircraft are already designed for flat-deck operations (except Harriers), every carrier aircraft would have to be redesigned to work with a catapult/ramp launch system. The inertia says stay with a flat deck.

@selectstriker2 Just the catapult area. Slanting the entire deck might just bring trouble to parked planes and landing.

@FreeMan I do think a curved "ski-jump" deck wouldn't be practical for engineering purpose, but perhaps a flat but non-level pitched up catapult wouldn't? If it is designed to be workable in rough sea, then a mere 5° pitch wouldn't hurt right?

So now, instead of the catapult shuttle having to negotiate a gentle slope, you're expecting it to make a sharp bend while under the massive load necessary to launch an aircraft? That's an even greater engineering challenge.

@FreeMan Why does it have to bend? Is it that hard to tilt the entire catapult? Is it because it might damage the keel? Maybe the support will reduce hanger space? Problem getting the aircraft in position?

Ah, so now the entire runway with the catapult is angled up? I suppose that eliminates the challenges I thought of, but the answers still stand - it's not necessary.

There's no benefit to a ski-jump catapult, and plenty of drawbacks. The purpose of a catapult is to rapidly accelerate an airplane up to flying speed in a short distance. Having an upward sloped angle makes it harder to get an airplane up to speed.
Angle of attack is going to vary aircraft to aircraft and be adjusted for performance factors such as weight, so any catapult ski-jump would have to be rapidly adjustable to allow for every launch, which would be complex.
There's just no point: once the airplane is up to flying speed its control surfaces will be used to adjust angle of attack, and if it isn't up to speed the crew will be reaching for ejection handles anyway, having a bit of a slope up at the end isn't going to help.

To answer your headline question:
A level deck allows for faster take-off with a greater payload than a ski-jump.
The ski-jump system has been used for decades, with the Royal Navy developing them in the 1970s. The Harrier "jump-jets" used in the Falklands in 1982 perfected the ski-jump launch - see this pic from Wikipedia's ski jump launch page for example:

The ski-jump is useful for reducing the take-off speed required to launch, but it has some risks for aircraft that cannot thrust vector as airspeed reduces after launch...
And that leads to other downsides, including reduced payload:

However, ski-jump launches cannot match the payloads made possible by high-speed catapult launches. While aircraft such as the F/A 18 that are normally catapult-launched can make use of a ski-ramp, this typically comes at the cost of a reduced capacity for either fuel or munitions, and thus negatively impacting mission scope significantly.

(also from that Wikipedia link)
So...why not use a catapult and a ski jump? Well, the catapult would have to stop before the ski jump, as the mechanical stresses involved in trying to have a straight then curved catapult make it impossible to maintain, so the highest speed would be at the base of the ramp. Which means the ramp then has to change the vector into an angled up course - that's a lot of strain on the undercarriage. So much so that even without a catapult, it broke various aircraft undercarriage when higher speeds were attempted.
Additionally, you don't want to expend energy to then have the speed reduce during the ramp phase - you get better bang for buck accelerating the length of the flat runway.

you're answering catapults vs ski-jump. I read the Op question as being, "why aren't catapults used together with ski-jumps?"

In re-reading the question I see what you mean. I don't think you could easily do it. You'd have to halt the catapult before the start of the ski jump, otherwise the engineering becomes horrible - will try and work out a useful edit to include that

How about non-curved "ski-jump"? I.e. flat runway angled pointing up? This option is not discussed in the answer.


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What are lateral boundaries?
In regard to:
103.17 OPERATIONS IN CERTAIN AIRSPACE No Person may operate an ultralight vehicle within Class A, Class B, Class C, or Class D airspace or within the lateral boundaries of the surface area of Class E airspace designated for an airport unless that person has prior authorization from the ATC facility having jurisdiction over that airspace.
What does lateral boundaries mean?
I see this question everywhere there are part 103 related discussions. Here's my take based on word etymology, which is how the legal definition will be derived.
Lateral comes from Latin "lateralis," meaning 'belonging to the side.' If you look at a map you see the horizontal boundaries of air space. Not the lateral boundaries. There are no sides depicted on a map, only the magenta boundaries right on the horizontal plane.
If you look at a 3D depiction, only then do you see the sides, or lateral boundaries of the airspace. The sides of the surface area of class e airspace extend from the ground up to where it joins regular class e. Part 103 liturally means the airspace between the sides. Those sides typically extend up 1200 ft above an airport where class e extends down to the ground, commonly referred to as the surface area of class e airspace. Above 1200ft it is no longer the surface area, it's just regular class e airspace above an airport.
So yes, you can fly in regular class e above the lateral boundaries of the surface area of class e as long as you are not between them, or within them, which is why the article says within the lateral boundaries. Within is synonymous with between, not above.

I think you are overthinking this…  Lateral just means the sides as you view the chart.

P.S. if you are looking for discourse this isn’t the place.  And I feel strongly that you shouldn’t answer the question in the question.  It’s totally OK to answer your own question, even encouraged.  Might even get enough upvotes to earn comment privileges.

Comments are not for extended discussion; this conversation has been moved to chat.

Just to reiterate-- if OP wishes to express a viewpoint on this subject he is more than welcome to post a new answer to the question designated as duplicate, https://aviation.stackexchange.com/q/74979/34686 -- plus it appears he now has enough rep to post to the chat as well. So -- let's discuss-- (in the chat) -- https://chat.stackexchange.com/rooms/142300/discussion-on-question-by-thatguyrj-what-are-lateral-boundaries -- I've posted some further thoughts there --


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Why do planes explode upon high velocity impacts?
I usually see fatal plane crashes end with a huge explosion, what causes the fuel to ignite?

It is possible worth clarifying that plans very rarely 'explode' since that has a specific meaning with respect to the burn front https://en.wikipedia.org/wiki/Deflagration. Aircraft instead normally have a rapid burnoff of fuel, potentially making a rising fire ball. This looks a lot like what films call 'an explosion' since burning fuel is cheap and impressive, but does change the precautions/prevention engineering. End result for the people on board is not meaningfully changed between 'explosion' and 'massive fireball' so does not impact the answer.

Hollywood special effects, maybe?

@PeterKämpf To be fair, most videos of airshow disasters do show planes deflagrating (instead of detonating) which most lay people would call "explode" even though it's just burning quickly instead of actually going "boom". The main cause is quite obvious - when you have lots of metal moving at high speed hitting other bits of metal moving at high speed you get a log of sparks which will cause fuel to burn

In the case of jet or turbo prop aircraft, the combustion section of the engine is continuously lit, and is not contained by a cylinder and valves like in a piston engine.  When fuel tanks rupture you have fuel and an open flame in close proximity that could ignite fumes.
And in a high speed impact, there is a chance that the engine case would also rupture, presenting an even greater opportunity for spilt fuel and an open flame to combine in a spectacular manner.


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How do I calculate the centre of gravity on the lateral (transverse) y-axis?
I have a real issue with determining the centre of gravity for the y axis of an aircraft, especially with selecting the reference/datum point and how to measure the displacement. Please I would really appreciate any form of help I can get.

Is this an academic exercise, or is there a real world reason why you need to do this?  Because in the real world it isn't important.  But if you have some reason it seems logical to me to use centerline as your datum.

To clarify my comment above in the context of John K' answer, i meant that it isn't important for loading cargo or passengers.  Fuel, however, is much more dense, and has a longer moment arm, so fuel imbalances will definitely affect roll control.

Transport airplanes have lateral C of G limits in terms of x distance from the centerline, but this only a certification requirement (related to roll authority in a critical case situation) and is only expressed for the flight crew as a fuel imbalance limit, fuel being the dominant factor, and is not accounted for in weight and balance data that the crew uses to determine if loading is correct.
The datum is the x=0 axis, with the C of G to the left, on, or right of it.  To calculate it, you would need an empty lateral C of G position, determined by weighing the airplane and working out the empty C of G offset to left or right of the x axis based on the weight difference between the left and right main gear weighing points, and the distance of those points from x.
Then you can add fuel and passengers and calculate the change if any to the lateral C of G. Since the passengers are normally randomly distributed on the left and right side of the cabin, the passengers can normally be considered neutral. Then it's only the effect of an unequal fuel load.
But as I said, for practical purposes, the fuel imbalance limits in the airplane's operating limitations account for the requirement to stay within maximum lateral C of G limits, and it's not necessary to calculate it.


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How realistic is atc-sim.com as an ATC simulator?
I was wondering how realistic https://atc-sim.com/ is.  I only play for fun, not for training, but I want to play as accurately as possible.

As Digital Dracula says, it's just a game.  The best way to experience something approaching the real deal is to join the VATSIM community. It's a virtual ATC system that integrates flight simulator players in a live dynamic ATC environment. It's much closer to the real thing because it's real people, some of whom are real controllers, and you can play it as a pilot using ATC services, or learn to be a controller yourself.
Be aware that the learning curve will be pretty steep compared to a stand-alone mental gymnastics game, because the participants make an effort to make things as realistic and true to life as possible, and as I said, a lot of real controllers participate and provide training tutorials and other resources.

VATSIM is very realistic (I am active there myself from time to time) when it comes to controller-pilot interaction. On the other hand, coordination (controller-controller), which is a huge part of real ATC, is almost not simulated at all, nor is interaction with airport staff, technicians, vehicles, bird control, the military, etc. etc. Just something to keep in mind :)

Do not forget IVAO. In Europe often more active.

@expeditedescent "VATSIM is very realistic" Until someone decides to launch a 747 straight up into the air like a space shuttle or something similarly ridiculous, if some of the Youtube videos I've seen of sessions of it are accurate.

@nick012000 Doing that is not allowed at these networks (IVAO, VATSIM), but they cannot put constraints on what your flight simulator does. They could only draw consequences from abusing behaviour later. Ignoring clearances and various kinds of trolling ar punishable and some mods are usually online and ready to respond.

@VladimirF Really? I'm thinking of videos like this one: https://www.youtube.com/watch?v=i_3pEOmLjJo Did anyone get banned for that?

@nick012000 The multiplayer in FSX is not the same as IVAO or VATSIM, these two networks are more governed by rules and also allow various kinds of simulators. One probably would not get banned for good but one might get thrown out for the day with a warning (depending on the further behaviour after being contacted by a mod). BTW, a "FUCKU" callsign would not be allowed either.

@nick012000 That's not VATSIM or IVAO, that's FSX multiplayer, which is far less realistic than the other two. Not too sure about IVAO, but VATSIM is very, very strict.

I've played atc-sim for fun myself some time ago, and it can be a nice way to pass time. However, in terms of realism, it only simulates a very narrow range of the tasks real controllers have to do. It gives you a (very basic) idea of what vectoring is like, and that's it. I'd say probably about 5% of what a real controller has to do is simulated, and given the text based interaction, even that part is not very realistic. So all in all, a fun game, but not at all realistic.

A chum of a chum who was an approach controller at CYUL told me that for the type of people with the mental aptitude and intelligence to make it through the training and do the job in the first place, with experienc your internal data processing becomes automated enough, plus the fact that the heavy workload periods come in relatively short rush hour waves of a couple hours each, the job was nowhere near the soul destroying pressure cooker that the popular media fostered.  She said her shift is about an hour and a half of heavy concentration, with a few hours of light work in between. True?

I agree, I wouldn't call the job stressful - at least not in the way it's often portrayed. Personally I find the quiet periods much more mentally draining, because I have to stay completely focused with very little to actually focus on. When it gets busy (up to a limit, of course) you sort of get "in the zone" and everything just runs like a well oiled machine. It's a great sense of accomplishment, and often surprise when your colleague comes to relieve you after what feels like 5 minutes but is often 1-2 hours

Long ago, back in the mid 80s, I tried out for Canadian ATC and took an initial screening test that was a booklet with 25 diagrams representing displays, with airplanes and tracks and data tables, and you had to answer two multiple-guess questions about potential conflicts for each diagram.  I didn't keep track of the time. I got to question 44 of 50, then "pencils down!!" I didn't answer the last 6.  I only got one wrong of the 44 I answered.  I should have just checked the last 6 off randomly and would have got one or two right, getting me over the 45/50 cutoff to move to the next stage.

authors: Vladimir F Героям слава


text:
What is that round antenna(?) on the bottom of a Piper Comanche?
Does anyone recognize this round object under the belly of a PA-24?

That is probably the ADF loop antenna. To read properly, it must be installed so the little airplane's nose points in the direction of flight.

I dont have the long sense antenna (at least not on the exterior) which makes sense, since I dont have an operable ADF any longer. And the little airplane is facing the right way.


text:
Is there any practical benefit of the Boeing-787 to perform near vertical takeoff?
In a promotional video for the Paris 2015 airshow, this video was released promoting the new Boeing 787's near vertical takeoff capability. Whilst highly impressive, I was wondering if there is any practical benefit or advantage of such an aircraft being able to perform this manouevre?
If there is no benefit of being able to climb at this angle of attack, then why was this aircraft designed with such a capability. But if the 787 wasn't designed to intentionally perform this, then does this mean other Boeing aircraft can do this? And if other aircraft can do this, then why hasn't this manouevre been marketed in such a way before?
In other words, why even bother to design an aircraft capable of this, if it serves no real benefit? And why market the aircraft as being able to do so in the first place?

Related: Could the Boeing 787-9 near vertical takeoff demonstration be performed by Airbus A350 aircraft?

The 787 was not designed to perform such a maneuver. Any jet airliner could do this.
What you need for such a steep climb is a good thrust to weight ratio (or enough excess speed to bleed off during the maneuver). A passenger twin-jet is designed such that it can still climb with at least 2.4% climb gradient after a single engine failure during a takeoff at the maximum takeoff weight (see e.g. this answer). When the aircraft carries no passengers and only enough fuel for a short demonstration flight, it is much lighter than that (source):

Aircraft
Operating Empty Weight
Maximum Takeoff Weight

Boeing 787-9
284,000 lb / 128,850 kg
560,000 lb / 254,011 kg

As you can see, the empty weight of a 787-9 is only about half its maximum takeoff weight. And if both engines are operating, that gives you a lot of excess performance for such a maneuver. At normal weights, the 787 could not perform such a steep climb and there is also no "practical benefit" in doing so since clearing terrain must still be possible after an engine failure, which results in much less climb performance.
Also note that the aircraft was not flying at a high angle of attack. It was flying at high pitch (although not nearly vertical). The angle of attack is the angle between the average chord line of the wings and the relative airflow. When climbing steeply, the angle of attack is therefore not necessarily high.

The 787 (and other commercial aircraft) are built to carry large loads, whether lots of cargo, or lots of passengers and bags.
So, when empty, their performance is much better.  If it can do a normal takeoff when full, then it can do a rocket-like takeoff when empty.
The near vertical takeoff is, indirectly, a demonstration of how much power it has when empty, and therefore, how much weight it can hold.  The maneuver itself is not useful.
However, its mostly just a visual spectacle.   Any commercial jet can do visually stunning  moves when not weighted down.   But it sure looks cool in front of an audience.


text:
How far apart are the rudder pedals on a Piper PA-28?
How far is the left rudder pedal from the right pedal on a Piper PA-28? I assume the left and right seats have equal pedal spacing, but if not, the left seat is what I would like to know.

Is "several inches" close enough?  What sort of accuracy or tolerance do you need?  (these things aren't typically listed in the POH, so it's an oddly specific request you will probably need an owner or renter to actually measure...)

Gosh I don't know off the top of my head, but I fly one and I'm going out for a bimble tomorrow, weather permitting, so I can measure if you need any sort of accuracy

@Jamiec just an approximation is fine; I'm building a set of pedals for my simulator.

Approximately how far apart your feet are if you are sitting with your legs out straight. ;)

What Chris said.  Most small GA airplanes have similar spacing.  Make it a comfortable fit for you, the actual measurement of a PA-28 really isn't important.  I'd almost be inclined to close the question based on the information not being useful to the general community.

@MichaelHall if you want to close it that's fine with me; I'm not concerned about it knowing that the precise separation won't matter.


text:
How precisely are pilots expected to be able to read altimeters?
I'm writing a program to practice reading three-hand analog altimeters. How precisely should a pilot be able to read one?

Not really an answer: many altimeters bounce around a fair amount. The engine makes a lot of vibration. Then, on the flip side, when there's no vibration the altimeter can stick. (On my glider, I have to tap on it to get it to respond for +-20' changes.)

As precisely as their vision allows?  As precisely as they are at reading an analog clock?

Most altimeters have tick marks every 20 feet, so a pilot should be able to read the altitude to 10 foot precision pretty easily.

Thank you! I made it go in 10-foot increments. https://kj7rrv.com/altimeter-practice/

@kj7rrv In an airplane with ambient vibration from an engine that breaks down internal stiction of the mechanism, the needle moves smoothly more or less like a sweep second electric clock in a continuous motion. You shouldn't make it move like a quartz clock; it's not realistic at all. You can easily discern 10 ft, half a tick mark. The required accuracy of the instrument itself is +/- 50ft but that doesn't apply here.

@kj7rrv Looks good. Might be a good idea to have it display some small negative altitudes to keep people on their toes.

@JohnK so you're suggesting that I make it move smoothly when it shows a new altitude?

@Chris-RegenerateResponse that's a good idea! Right now it goes 0 to 18000 feet. Maybe change it to something like -300 to 18000?

@kj7rrv Apparently the lowest airport in the world (LLMZ) is at elevation -1266, so perhaps -1260 or -1270 would be a fun minimum to use?

@kj7rrv yes the needle is being driven by a bellows expanding and contracting with air pressure that drives the needle through a gear train. So the movement is progressive in proportion to the pressure change. Internal friction and age can make them move in little fits and starts, but a new one, especially if there is any vibration to shake the mechanism that brakes down internal static friction, will move smoothly. So assume it moves like an old fashioned electric clock sweep second hand.

@JohnK For an app meant to practice reading altimeters at a glance it's probably better if it doesn't move realistically. If it does you'd be able to tell the rough altitude just from how long it takes without worrying about even reading the needles.

Not quite sure what you mean, but perhaps.

Altimeter reading precision is not an item on the Airman Certification Standards - formerly known as Practical Test Standards (PTS) - used by the FAA as the grading criteria for checkrides. The ability to hold altitude within certain tolerances during various maneuvers is.
The private pilot (PPL) criteria are +/- 200 feet in cruise and +/- 100 feet in the traffic pattern and most maneuvers. So you need to be able to read the altimeter well enough to judge that.
In practice, as the other answer stated, it's pretty easy to read within 10 or 20 feet on all types of "steam gauge" (moving needle) and digital flight displays I've seen.

Thank you! I made it go in 10-foot increments. https://kj7rrv.com/altimeter-practice/


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What are those ball-shaped things on some helicopter blades?
I saw these things swinging in a video of a rotor from the blade point of view. I don't know what are they for but I am quite curious. And why some helicopters have it and some don't?

Source: alamy

Source: Helicopter Main Rotor Blade In Flight Slow Motion -- YouTube

Reminder: please no answers (or attempts) in the comments

The devices are pendulum (or pendular) absorbers:
Vibration suppression of helicopter blades by pendulum absorbers - Imao Nagasaka.
The pendulum absorbers are used for suppressing the vibrations in helicopter blades. Compared to fixed weight attached to the blade, the pendulum configuration allows for shifting of the damping range depending on rotor angular speed.

@ymb1 My educated guess would be that you need damping during the start-up sequence, too, to prevent harmful oscillations when the rotor RPM and blade eigenfrequency (which changes due to centrifugal stiffening) coinciding.

Good point @ymb1! I'm guessing Sanchises  has it right.

The hinged arrangemet may also be better, since the mass is connected to the rotor by a shaft, thus affecting which frequency modes it dampens (not sure about terminology here). A simple fixed mass would have mechanically very different characteristics.

@Jpe61 Yes. The hinge decouples the mass's moment of inertia in the axis going across the blade - it's as if the mass had zero moment-of-inertia in that axis. I'm not sure whether this effect is necessary/useful, but it sure is there.


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Cause of Drag From Windmilling
I know that when a propeller is windmilling it creates drag, I assume this is from the propeller being flat against the air causes drag, but propellers are designed to essentially pull an airplane forwards with rotation so wouldn't the incoming air spin the propeller and create thrust? Does the wind not cause the propeller to spin fast enough for thrust and only just wastes energy by using it up spinning a propeller too slow to actually help with propulsion?

a windmilling prop is extracting work from the air moving through the prop disc, and using that work to rotate the dead engine. the power extracted is approximately equal to the speed of the air times the drag force exerted on the prop.
This is the reverse of the situation in which a running engine is spinning a propeller which performs work on the air, generating a reaction force which is expressed as thrust.

So it's only the friction in the engine?

@Bergi: plus the aerodynamic drag of the blades

@Bergi, "friction" is probably the wrong word.   Most of the work goes into compressive heating of the air in the cylinders.

@BowlOfRed Virtually all of the work that goes into compressing the air should be recovered back during the power stroke (even when there is no combustion); so there shouldn't be a net energy loss. What actually absorb the energy (from the windmilling prop) are the pumping losses, i.e. Induction and exhaust stroke. Induction stroke: This is because the engine is essentially pulling a vaccum through the induction manifold (a volumetric efficiency of less than 10% would be typical with the throttle fully closed). This effect can be greatly reduced by fully opening the throttle valve.

Secondly, exhaust stroke: This loss results from the back pressure in the exhaust manifold, and the piston trying to force the exhaust gases against it. For example, a muffler alone can cause around 10 kPa (1.5psi) of back pressure. Apart from the pumping losses, the second biggest power absorber is the internal friction of the engine.

@AdityaSharma, "Virtually all of the work that goes into compressing the air should be recovered back during the power stroke".  That sounds good in theory, but if it were actually true engines wouldn't be so difficult to turn over by hand.  There is a huge difference in the effort needed to turn an engine over when you remove the spark plugs to take away the compression.

@AdityaSharma: you should write it in an answer

@MichaelHall I invite you to this chat room to continue this discussion

Normally, an airplane converts thermal power (from combustion) into thrust (mechanical power).  When the propeller is windmilling, drag (also mechanical power) is converted to thermal power.
As mentioned in another answer, windmilling the propeller rotates the engine.  The engine moves gasses around and it has internal friction.  Therefore the engine consumes power which contributes to the power of drag.

Does the wind not cause the propeller to spin fast enough for thrust...?

Think of it in terms of cause and effect.  If you had a windmill but no wind, nothing is moving, there is no cause or effect.

If the wind starts to blow, the windmill will turn.  The wind is the cause, the windmill turning is the effect.
If you instead hooked a motor up to the windmill and forced it to turn, the air would start to move.  In this case, the windmill turning is the cause, and the air moving is the effect.

It doesn't make sense to think that in the first case the wind causing the windmill to turn is what causes the wind.  That just isn't how things work, energy must be transferred from one form into another - it isn't just created.
An airplane propeller is either being powered by the engine, moving air and producing thrust, or it is unpowered, and being rotated by the force of the air.  It can't do both.  When it is unpowered it is producing drag, not thrust.  The speed at which the wind rotates an unpowered propeller is irrelevant to whether or not it is producing thrust.
I would encourage you to do some research into Newton's laws of motion for more background on this.